Constrained motion

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[Mentors note: Thread moved from the Classical Physics forum after it had been replied to, hence the lack of a homework template]

Hey I have been trying to solve this problem for the last few days with no luck.

Question:

a car starts from rest at the origin and travels along the path given by y=0.2 x ^(3/2) and picking up speed in accordance with v = 4.8 t. At which x position does the car skid off the path? Friction force is known as well (Ff)


Attempt at answer:

I parametrized the curve in terms of time, r(t) = <t, 0.2t^3/2> , calculated the radius of curvature (R) as a function of t, and used this to solve for the centripetal force using Fc = mv^2 / R , where v=4.8t . Then I said that the force of friction always acts in the inward normal direction to keep the car from going off the path. The road exerts a reaction force on the vehicle equal to - Fc . Therefore using newtons second law i solved for the position t when Fc = Ff. This all makes sense to me although Ive been told this is not the correct method. Any thoughts are appreciated. Thanks
 
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Answers and Replies

  • #2
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y is some distance and v is speed? So we know y-distance as function of time and we know speed? Does that mean you have to find the x-position based on those two equations?

Or is it just a typo and the equation for y depends on x?
 
  • #3
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Ah yea my bad, the path is y(x) = 0.2 x^3/2 and the speed is a function of time
 
  • #4
Svein
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Since v(t) is not given as a vector, we must assume that it is the speed along the road. Now the arc length element ds is given by ds2=dx2+dy2. Your equations gives y(x) = 0.2 x^3/2, so dy = 0.2⋅3⋅x2/2 dx or [itex]ds=\sqrt{1+0.3x^{2}}dx [/itex]. Since v(t) = at, where a = 4.8m/s2, the position at t = τ is given by [itex] s(\tau)=\frac{1}{2}a\tau^{2}[/itex]. Now, the tangent vector [itex]\vec{t}= \frac{d\vec{r}}{ds} [/itex] and the curvature is given as [itex]\kappa = \vert\frac{d\vec{t}}{ds}\vert [/itex]. ...
 
  • #5
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Together with the centripetal force, we also have to take the acceleration along the direction of motion into account. The friction force has to "cover" both.
 
  • #6
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Thanks for the replies. Still a little confused.

The question is asking for the x position of the vehicle, not the distance covered along the path so would I still need to use arc length?

I think what im getting confused about is direction of the friction force. Am I wrong to assume that the friction force only acts in the direction normal to the path of travel? Would there be a force that is always acting in the tangential direction Ft=4.8m since the velocity is increasing at a constant rate? Then the net force would be the tangential force plus the centripetal force and friction would have to act to oppose those combined forces?

Oh also note i corrected the equation of the path to make it more clear. Its y(x) = 0.2 x ^(3/2)
 
  • #7
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You'll need the arc length to find the relations y(t) and x(t), otherwise how would you know where you are at a specific speed?
Then the net force would be the tangential force plus the centripetal force and friction would have to act to oppose those combined forces?
Right.
 
  • #8
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Hmm okay. I have y(t) and x(t) now so I have r(t) = < x(t), y(t) > . Only thing is that these vectors are getting really complicated to the point i can only work with them in maple, and most of my class has never even seen vector calculus so im starting to think there has to be a much simpler solution.

since the centripetal force is given by Fc= (mv^2) /r , and v is known, would it be possible to solve it using just this info? like when the magnatude of tangental and centripetal force components is equal to force of friction? So far im just getting more confused.
 
  • #9
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You'll need r for that formula.
I think the curvature approach is a detour. x(t) and y(t) allow to calculate the acceleration in y and x direction, which gives the total acceleration.
 
  • #10
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I think i might have made a mistake getting x(t) and y(t).

I integrated ds from 0 to x so I had arc length as function of x, s(x) and then set s(x) = 1/2 *4.8*t^2 and solved for x. as far as I know this would be x(t) which i could use in my parametrization of the curve r(t) but this function is impossible to work it takes up a full line in maple and is full of square roots and such. Seeing as though we are not expected to know vector calc, there must be another solution method that I am missing. Even when i continue with maple to solve the problem I end get getting 4 solutions most of which are complex numbers.
 
  • #11
Svein
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Oh also note i corrected the equation of the path to make it more clear. Its y(x) = 0.2 x ^(3/2)
This changes several terms. Now dy = 0.2⋅(3/2)⋅x^(1/2)⋅dx, and [itex]ds=\sqrt{1+0.09x}dx [/itex]...
 
  • #12
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Ah yea I realized that before and made the correction but thanks. What would x(t) be? I have an answer but its so long im having a hard time thinking its right.
 
  • #13
Svein
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What would x(t) be?
Taking the indefinite integral on each side, we get [itex] s=\frac{2}{3\cdot 0.09}(\sqrt{1+0.09x})^{3}+C[/itex]. SInce x=0 when s=0, C=0.
 
  • #14
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So then i set this equal to 1/2 * 4.8 ^ t^2 and solved so I have x(t). so now I have my position vector in terms of time. r(t)=<x(t) , 0.2*x(t)^(3/2) > , now differentiating this x2 i get a(t) i believe. Since the only force acting on the vehicle is friction, and this opposes the net acceleration, would the vehicle slip when the magnitude of a(t) = Ff ?
 
  • #16
Svein
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now differentiating this x2 i get a(t) i believe
Not quite. [itex]\frac{d\vec{r}}{dt}=\vec{v}(t)=(\frac{dx(t)}{dt},\frac{dy(t)}{dt}) [/itex]. The acceleration is then given by [itex]\frac{d\vec{v}}{dt}=\vec{a}(t)=(\frac{d^{2}x(t)}{dt^{2}}, \frac{d^{2}y(t)}{dt^{2}}) [/itex]. The magnitude of the acceleration is then √(ax(t)2+(ay(t)2).
 
  • #17
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Thanks for help, finally got the correct answer last night
 

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